Question

find the derivative of $f(x)=sqrt{x^{2}+sqrt{x^{3}+cos(x^{2})}}$.
$f(x)=\frac{1}{2}x^{2}+sqrt{x^{3}+cos(x^{2})}^{-\frac{1}{2}}2x+\frac{1}{2}x^{3}+cos(x^{2})^{-\frac{1}{2}}3x^{2}-sin(x^{2})2x$
the function is not differentiable
$f(x)=\frac{1}{2}x^{2}+sqrt{x^{3}+cos(x^{2})}^{-\frac{1}{2}}2x+\frac{1}{2}3x^{2}-sin(x^{2})2x^{-\frac{1}{2}}$
$f(x)=\frac{1}{2}x^{2}+sqrt{x^{3}+cos(x^{2})}^{-\frac{1}{2}}2x+\frac{1}{2}x^{3}+cos(x^{2})^{-\frac{1}{2}}$
$f(x)=\frac{1}{2}x^{2}+(x^{3}+cos(x^{2}))^{\frac{1}{2}}^{-\frac{1}{2}}2x+\frac{1}{2}x^{3}+cos(x^{2})^{-\frac{1}{2}}3x^{2}-sin(x^{2})2x$

Explanation:

Step1: Let \(u = x^{2}+\sqrt{x^{3}+\cos(x^{2})}\), then \(f(x)=\sqrt{u}=u^{\frac{1}{2}}\)

By the chain - rule, \(f^{\prime}(x)=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^{\prime}\)

Step2: Find \(u^{\prime}\)

Let \(v = x^{3}+\cos(x^{2})\), then \(u=x^{2}+\sqrt{v}=x^{2}+v^{\frac{1}{2}}\)
\(u^{\prime}=2x+\frac{1}{2}v^{-\frac{1}{2}}\cdot v^{\prime}\)

Step3: Find \(v^{\prime}\)

\(v = x^{3}+\cos(x^{2})\), so \(v^{\prime}=3x^{2}-\sin(x^{2})\cdot2x\)

Step4: Substitute \(v^{\prime}\) into \(u^{\prime}\) and then \(u^{\prime}\) into \(f^{\prime}(x)\)

\(u^{\prime}=2x+\frac{1}{2}(x^{3}+\cos(x^{2}))^{-\frac{1}{2}}(3x^{2}-2x\sin(x^{2}))\)
\(f^{\prime}(x)=\frac{1}{2}[x^{2}+\sqrt{x^{3}+\cos(x^{2})}]^{-\frac{1}{2}}[2x+\frac{1}{2}(x^{3}+\cos(x^{2}))^{-\frac{1}{2}}(3x^{2}-2x\sin(x^{2}))]\)

Answer:

\(f^{\prime}(x)=\frac{1}{2}[x^{2}+\sqrt{x^{3}+\cos(x^{2})}]^{-\frac{1}{2}}[2x+\frac{1}{2}(x^{3}+\cos(x^{2}))^{-\frac{1}{2}}(3x^{2}-2x\sin(x^{2}))]\) (First option)